How do you solve ?

#(1-cos8x)/(1+tgx) = 0#

Correct answer is #x=(kπ)/2# or #x=π/4 +kπ#

1 Answer
Nghi N.
Jun 24, 2018

#x = (kpi)/4#,
except the 2 values: x = (-pi/4), and x = (3pi)/4

Explanation:

#(1 - cos 8x)/(1 + tan x) = 0#
#(1 - cos 8x) = 0# (1)
Condition #tan x != -1# , or #x != - pi/4#, and #x != +(3pi)/4#.
Solve equation (1):
#cos 8x = 1#
Unit circle gives:
#8x = 2kpi# -->
#x = (kpi)/4#
(except when #x = - pi/4# and #x = (3pi)/4#)
Specifically,
When k = -1, #x = -pi/4# (rejected)
When k = 3 --> #x = (3pi)/4# (rejected)
When k = 7 --> x = (7pi)/4 = - pi/4 (rejected)

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